Examples of proofs of irrationality without using contradiction

Question

I'm just wondering if there exists proofs that certain numbers are irrational that do not begin by saying some like along the lines of "assume $k=a/b$ for integers $a$ and $b$" and then deduce a contradiction. My question is, are there any irrationality proofs out there that somehow prove irrationality directly or do they all somehow utilize proof by contradiction?

Answer 1

Comment: I remember coming across a real neat generalized way of proving the irrationality of $\sqrt{a}$ in the American Mathematical Monthly (credit due to Amrik Singh Nimbran). Maybe the following will be something along the lines of what you are looking for (Daniel Escudero's comment makes a good point though):

Generalized proof: Let $a$ be a square-free positive integer and $[\sqrt{a}]=b$. where $[\;]$ denotes the greatest integer contained in the square root. Then $b^2<(\sqrt{a})^2<(b+1)^2$. Thus, $a$ lies between two consecutive squares; hence, $\sqrt{a}$ is not an integer. Suppose that $\sqrt{a}$ is rational, i.e., $\sqrt{a}=\frac{m}{n}, 1<n<m<(b+1)n; m,n\in\mathbb{N}$. Using a division algorithm, we may write the following: $m=bn+r, n>r, r\in\mathbb{N}$. Hence, $\sqrt{a}=\frac{bn+r}{n}$. Let $g=(n,r)$. So, $n=gs$ and $r=gt, s>t, (s,t)=1; g,s,t\in\mathbb{N}$. Then $\sqrt{a}=\frac{bs+t}{s}$. Hence $$ (a-b^2)s^2=t(2bs+t).\tag{1} $$ Since $t\mid(t(2bs+t))$ and $(s,t)=1$, either $(a-b^2)=tu, u\in\mathbb{N}$, or $t=1$. In the first case, we would have $s(us-2b)=t\Rightarrow s\mid t$, which is impossible for $s>t$. In the second case, that is, $t=1$, we obtain $$ s\{(a-b^2)s-2b\}=1.\tag{2} $$ Equation $(4)\Rightarrow s\mid 1\Rightarrow s=1$, which contradicts our supposition that $s>t=1$. Further, it leads to $a=(b+1)^2$, contradicting our supposition $a<(b+1)^2$. Hence, $\sqrt{a}$ is irrational. $\blacksquare$

Answer 2

Consider the polynomial $x^2-2$, by Eisenstein's criterion this polynomial is irreducible over $\mathbb{Q}$. It follows that the extension degree of $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$ is $2$, hence strict. This shows that $\sqrt{2}\notin \mathbb{Q}$.

You can go through the proofs of Eisenstein's criterion and the other theory involved, we nowhere use that $\sqrt{2}$ is irrational.